Equation for the oxidation reaction for KMnO4 solution

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Discussion Overview

The discussion revolves around the oxidation reactions involving KMnO4 solutions, particularly in the presence of dilute HCl and other compounds. Participants explore half-equations, oxidation states, and the balancing of redox reactions, addressing both theoretical and practical aspects of these chemical processes.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions what is being oxidized in the reaction involving KMnO4 and dilute HCl.
  • Another participant suggests that chlorides are oxidized, prompting further inquiry into the oxidation states of chlorine.
  • There is a discussion about the oxidation half-equation for chloride ions, with some proposing Cl- to Cl2 as the product.
  • Participants debate the reduction half-equation for permanganate, with one stating it involves MnO4- and hydrogen ions.
  • Concerns are raised about balancing the overall reaction, with suggestions to adjust coefficients to ensure electron cancellation.
  • In a later part of the discussion, participants explore the oxidation states of sulfur in Na2SO3, with some asserting that sulfur is oxidized to a +6 state.
  • There is confusion regarding the oxidation states and the role of spectator ions in the reactions discussed.
  • Participants also discuss the oxidation of hydrogen peroxide in a reaction with KMnO4, with differing views on the products formed.
  • One participant mentions the potential oxidation of Mn2+ to MnO4- and questions the half-equation for this process.

Areas of Agreement / Disagreement

Participants express various viewpoints on the oxidation processes and the corresponding half-equations, indicating that multiple competing views remain. The discussion does not reach a consensus on several points, particularly regarding oxidation states and the roles of different species in the reactions.

Contextual Notes

Participants highlight the importance of balancing half-reactions and the potential complexity of oxidation states, noting that some assumptions about oxidation states may not apply universally. There is also mention of the need for clarity in defining the roles of ions in the reactions.

agrocadabra
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Ok so the question goes as such: Write the half equation for the oxidation reaction and then the full net ionic equation for KMnO4 solution mixed with dilute HCL. My main problem is what is getting oxidised?
 
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Chlorides.

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methods
 


chlorides are getting oxidised? how so?
 


It is just a matter of potentials, permanganate in solutions of pH low enough is a very strong oxidiser.

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ok so in this case.. the chlorine will have an oxidation state of minus one to begin with and an oxidation state of -2 to end with?
 


What is oxidation?

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sorry its a loss of electrons.. so the Cl- must lose and electron to become Cl?
 


so the oxidation half equation would be Cl- ---> Cl + e ?
 
  • #10


why Cl2 as opposed to Cl? So it oxidises chlorides to chlorine.. why?
 
  • #11


So the oxidation half equation would be:
2Cl- --> Cl2 + 2e ?
 
  • #12


Which gases are diatomic?

Edit: yes to your oxidation half reaction.

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  • #13


But the reduction equation will be: MnO4- + 8H+ + 5e ---> Mn2+ + 4H2O ... when the oxidation and reduction equations are added together the electrons won't cancel..
 
  • #15


uhh so multiply all the left side by 2 and the right by 5? then 10 electrons on each and they will cancel?
 
  • #16


oh no not that easy.. sorry.. i shall read over that link.. thankyou sooo much foir your help so far.. absolute champion!
 
  • #17


agrocadabra said:
uhh so multiply all the left side by 2 and the right by 5? then 10 electrons on each and they will cancel?

That will probably do - it may happen that coefficients in the reaction after will be not the smallest ones, but at least reaction would be correctly balanced.

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  • #18


Ok so I get..

2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ---> 2Mn2+(aq) + 8H2O(l) + 10Cl2(g)

I've got three more of these to do.. ill be up all night.. argh!
 
  • #19


agrocadabra said:
2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) ---> 2Mn2+(aq) + 8H2O(l) + 10Cl2(g)

Divide everything by 2.

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  • #20


Ok so the next one is KMnO4 and Na2SO3 in the presence of dilute H2SO4.. however what the heck is the oxidation state of Na2SO3?
 
  • #21


The oxidation state of the SO3 should be 0.. but what about the Na2? Would the Na2 be assigned a negative oxidation state because of its greater affinity for electrions? so Na is usually + would the oxidation state of Na2 in this case be -2?
 
  • #22


Don't think in terms of ON, if you are going to balance thorugh half reactions.

Besides, there is no such thing as oxidation state of Na2SO3 - oxidation state is assigned to individual atoms, not molecules.

Na2SO3 is just a salt - it dissociates giving 2Na+ and SO32-. Na+ are just spectators.

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methods
 
  • #23


so the Na and K ions are spectators in this one?
 
  • #24


Yes.

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methods
 
  • #25


So sulphur is going to be oxidised.. what is the sulfur trioxide's formal charge? is it -2 or 0? if it is -2 then the sulfur has a ON of +4 how the heck do I know what its oxidation number goes to?
 
  • #26


is the sulfur oxidised to s+6 ions?
 
Last edited:
  • #27


There is no trioxide here, I have alredy explained what is the ion that will be oxidised.

Yes, +6 will be the final ON of sulfur. No idea why you are using ON - they are not necessary when balancing through half reactions.

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methods
 
  • #28


it seems weird that the sulfur would be oxidised... would it be that the SO3 ion is a spectator as well and that the hydrogen ions are oxidised to hydrogen gas?
 
  • #29


Why is it +6, why not something else, something higher? It just seems so unconcrete.. I'm using ON numbers because I don't know how else to go about it.
 
  • #30


So it would be S+4 ---> S+6 + 2e?
 

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